Edited Invalid date
8.7 Aldol Reactions
In this chapter, we've learned how to make enolates, and we've used them to attack various electrophiles.
We will look at what happens when an enolate attacks a ketone or aldehyde.
Suppose we use hydroxide as a base and start with a simple ketone.
The enolate can attack the electrophile if we do this.
The equilibrium will replenish the supply.
It turns out that there is more than one person present.
There is a lot of ketone present and the enolate is in equilibrium with it.
When ketones are attacked, we devoted an entire chapter to the reactions that take place.
An alkoxide intermediate is formed when the enolate attacks the ketone.
Our golden rule is to try and re-form the carbonyl group, but not expel H- or C-.
We can't leave groups that can be expelled.
There is only one way to remove the charge.
This is the beginning of the reaction.
Whenever an enolate attacks a carbonyl group, the structure of the enolate will always be the same.
The carbonyl group of the ketone is being attacked by the alpha carbon of the enolate.
The OH group will always be in the alpha position.
The reaction doesn't stop at the b-hydroxy ketone.
The product has a double bond with the carbonyl group.
Between the a and b positions is where the double bond is located.
We can control how far the reaction goes in the laboratory.
Carefully controlling the conditions of the reaction.
You should know the proper terminology.
A small molecule is liberated in a condensation reaction.
The small molecule can be either N or CO.
We can't call it a condensation reaction if a water molecule is not lost in the process.
We call it an aldol addition.
The mechanism of aldol condensation is easy to understand.
It can be difficult to see what reagents to use in a synthesis.
Two alpha protons and an oxygen atom are being removed from one ketone.
The curved arrows and intermediates are shown in a mechanism.
This way of thinking about the reaction might be useful when proposing a new drug.
We will get practice with the mechanism to get the answer.
We need to make sure that we can use our simple method for drawing the expected product.
It's an efficient way to draw the product.
If there is a condensation, draw the a,b-unsaturated ketone that is produced.
One molecule of ketone was deprotonated to give an enolate, which then attacked another molecule of the same ketone.
The ketones are different from each other.
Care must be taken to avoid generating many different products.
You can't really tell which ketone will be converted into the enolate.
There will be more than one type of enolate and more than one type of ketone present.
There are a number of possible reactions that can happen, and this will result in a mixture of products.
It is important to avoid these types of situations.
There is a very easy way to avoid this issue.
If there is no alpha protons in one of the ketones, it cannot form an enolate.
The compound has no alpha protons.
It can't be converted into anolate.
One way to cross aldol is to make sure that one of the reagents has no alpha protons.
The number of potential products will be minimized.
If both starting ketones have alpha protons, your textbook may or may not show methods for crossing aldol.
If you are responsible for such methods, you should look through your textbook and lecture notes.
The method we used before can be used again.
We need to do it in a different way.
We break the molecule apart to get water.
We have to decide which fragment has the oxygen atom and which fragment has the protons.
The fragment on the left has a carbonyl group, so it must get the two alpha protons.
Review flashcards and saved quizzes
Getting your flashcards
Privacy & Terms